如何将两行的值合并到一个“重复 ID”中

Question

我试图找到答案，但我发现的任何东西都不适用于我的情况，因此我将不胜感激：

场景：我的客户在一个表中具有 ID 和个人信息，然后我在另一个表中指示属于这些 ID 的注释。 当我加入表格时，我会在多行中得到多个注释。如下图。

T1. ID -T1. Name    -T1. Birthday   -T2. Note                  -Row number
2      -Peter       -11/20/1990     -deciding                  -1
2      -Peter       -11/20/1990     -purchased                 -2
3      -David       -12/22/1962     -Presentation scheduled    -1  
3      -David       -12/22/1962     -Presentation completed    -2
3      -David       -12/22/1962     -Purchased                 -3
4      -Anna        -5/07/1992      -Ignored                   -1

我一直在徘徊的是如何在第一行的末尾添加注释 2 和 3 并消除重复项？

T1. ID -T1. Name    -T1. Birthday   -T2. Note               -Note_2      -Note_3
2      -Peter       -11/20/1990     -deciding               -purchased  
3      -David       -12/22/1962     -Presentation scheduled -Presentation -completed    Purchased
4      -Anna            -5/07/1992  -Ignored                -Null        -Null 

您可以使用此查询来查看表格：

declare  @t1  table ( ID int not null,
                      Name varchar (50),
                      birthday date)

declare @t2 table (ID int not null,
                    Note nvarchar(max))


insert into @t1 values ( 2 , 'Peter' , '11/20/1990')
insert into @t1 values ( 3 , 'David' ,'12/22/1962')
insert into @t1 values ( 4 , 'Anna' , '5/07/1992')

insert into @t2 values (2 , 'deciding')
insert into @t2 values (2 ,'purchased')
insert into @t2 values (3 ,'Presentation scheduled')
insert into @t2 values (3 ,'Presentation completed')
insert into @t2 values (3 ,'Purchased')
insert into @t2 values (4 ,'Ignored')



SELECT * 
        , ROW_NUMBER() OVER (PARTITION BY t1.ID ORDER BY t2.id) AS 'ROW NUMBER'
FROM @t1 as t1
left outer join @t2 AS t2 ON t1.ID = t2.ID 

Answer 1

这是一个完整的脚本。我知道它可能不如上面那个简洁，但它不需要任何额外的附加组件就可以工作，而且也很容易解释。

我唯一不确定的是您是否需要三个单独的注释列或一列包含串联注释。我使用了连接注释方法......

declare @t1 table
(
    ID int not null,
    Name varchar(50),
    birthday date
)
declare @t2 table
(
    ID int not null,
    Note nvarchar(max)
)
insert into @t1
values
(2, 'Peter', '11/20/1990')
insert into @t1
values
(3, 'David', '12/22/1962')
insert into @t1
values
(4, 'Anna', '5/07/1992')
insert into @t2
values
(2, 'deciding')
insert into @t2
values
(2, 'purchased')
insert into @t2
values
(3, 'Presentation scheduled')
insert into @t2
values
(3, 'Presentation completed')
insert into @t2
values
(3, 'Purchased')
insert into @t2
values
(4, 'Ignored')

--Create table of all notes and what person they belong to.
declare @Notes table
(
    NoteID int not null,
    ID int not null,
    Note nvarchar(max)
)

--Insert into note table
insert into @Notes
(
    NoteID,
    ID,
    Note
)
SELECT ROW_NUMBER() OVER (ORDER BY t2.note),
       t1.ID,
       Note
FROM @t1 as t1
    left outer join @t2 AS t2
        ON t1.ID = t2.ID

--Create a table of person ID and concatenated notes.
declare @ConcatNotes table
(
    ID int not null,
    Note nvarchar(max)
)

--Insert just people first
insert into @ConcatNotes
(
    ID
)
select distinct
    ID
FROM @Notes

declare @NoteCount int
declare @i int = 1

select @NoteCount = max(noteid)
from @Notes

--While loop to loop through ALL notes by person ID and concatenate with semicolon. Change semicolon to whatever you want.
while (@i <= @NoteCount)
BEGIN
    update a
    set Note = isnull(a.Note + ';', '') + b.Note
    --select *
    from @ConcatNotes a
        inner join @Notes b
            on a.id = b.ID
    where b.Noteid = @i
    set @i = @i + 1
END

--A select to show the person ID and the concatenated notes.
select *
from @ConcatNotes

--The final query you were looking for.
select a.*,
       b.Note
from @t1 a
    inner join @ConcatNotes b
        on a.ID = b.ID

Answer 2

我在这里找到了这个 CLR 解决方案： https://groupconcat.codeplex.com 并成功使用。

基本上，您的查询将变为：

;WITH example AS (
SELECT t1.id, t1.Name, t1.birthday, t2.id AS t2id, t2.Note 
        , ROW_NUMBER() OVER (PARTITION BY t1.ID ORDER BY t2.id) AS 'ROW NUMBER'
FROM @t1 as t1
left outer join @t2 AS t2 ON t1.ID = t2.ID 
)
select ID, name, birthday, [dbo].[GROUP_CONCAT_D]( Note, '   -') AS Notes
from example
group by ID, name, birthday

结果如下所示：

ID  name    birthday    Notes
2   Peter   1990-11-20  deciding   -purchased
3   David   1962-12-22  Presentation scheduled   -Presentation completed   -Purchased
4   Anna    1992-05-07  Ignored